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cascade amplifier with common emitter amp and common collector amp

majdi

Jul 10, 2010
33
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33
main.png


I need to find IE1, IE2, re1, re2 and overal voltage gain.

First i should open all capacitor first right?. it should be like this:
main2.png


How possible i can calculate since there dont have enough value in this circuit? like Ic or Ib.
 

Harald Kapp

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Start by noting that for each transistor IE=(1+beta)*Ib
Set up the equations applying Kirchhoff's laws and solve for the operating point.
 

majdi

Jul 10, 2010
33
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Here my final draw:
main3.png


IC1 = IB2

IC1 = VR1/R1

VR1 = Vcc - (Vceq1+VRE)

----------------------------------------------
or

Vceq2 = Vcc - (VR4 + VR5)

I should find IB2 first..let me think........
 

Harald Kapp

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IC1 <> IB1
IC1 flows through the collector of Q1, not into the base of Q2.
IC1+IB2 = VR1/R1
...
 

majdi

Jul 10, 2010
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IE1 = (VCC – VCE) / (R1 + R2 + R3)
= 12v – 0v / 20k + 2k + 8k
= 12v / 30k
= 0.4mA

Let assume VCE Q1 = 6.0v

IE1 = (VCC – VCE) / (R1 + R2 + R3)
= 12v – 6v / 20k + 2k + 8k
= 6v / 30k
= 0.2mA

VE Q1 = IE1 x (R2+R3)
= 0.2mA x 10k
= 2v

VB Q1 = VE1 + VBE1
= 2v + 0.7v
= 2.7v

VB Q2 = VCC – (R1 x IE1)
= 12v – 4v
= 8v

VE Q2 = VB Q2 – VBE Q2
= 8v – 0.7v
=7.3v

VR5 = ½ x VE Q2
= 3.65v

*VR5 should be near to VB Q1 = 2.7v, this show IE1 is too low, we need to increase it. Try IE1 = 0.3mA

VE Q1 = IE1 x (R2+R3)
= 0.3mA x 10k
= 3v

VB Q1 = VE1 + VBE1
= 3v + 0.7v
= 3.7v

VB Q2 = VCC – (R1 x IE1)
= 12v – 6v
= 6v

VE Q2 = VB Q2 – VBE Q2
= 6v – 0.7v
=5.3v

VR5 = ½ x VE Q2
= 2.65v

*VR5 should be near to VB Q1 = 2.7v, this show equal to VB Q1 to low. Try with IE1 = 0.25mA

VE Q1 = IE1 x (R2+R3)
= 0.25mA x 10k
= 2.5v

VB Q1 = VE1 + VBE1
= 2.5v + 0.7v
= 3.2v

VB Q2 = VCC – (R1 x IE1)
= 12v – 5v
= 7v

VE Q2 = VB Q2 – VBE Q2
= 7v – 0.7v
=6.3v

VR5 = ½ x VE Q2
= 3.15v

*This result is close for VB Q1.


Let IE1 = 0.24mA

VE Q1 = IE1 x (R2+R3)
= 0.24mA x 10k
= 2.4v

VB Q1 = VE Q1 + VBE Q1
= 2.4v + 0.7v
= 3.1v

VC Q1 = VCC – (IE1 x R1)
= 12v – 4.8v
= 7.2v

VB Q2 = VC Q1
= 7.2v

VE Q3 = VB Q2 – VBE Q2
= 7.2v – 0.7v
= 6.5v

IE2 = VE Q2 / (R4 + R5)
= 6.3v / 2k
= 3.15mA


So: IE1 = 0.25mA and IE2 = 3.15mA.

re1 = VT / IE1
= 26mV / 0.25mA
= 104 Ohm.

re2 = VT / IE2
= 26mV / 3.15 mA
= 8.25 Ohm.


IB1 = IE1 / Beta
= 0.25mA / 100
= 0.0025mA


IC1 = 1+Beta x IB1
= 101 x 0.0025
= 0.253mA

Av1 = IC1xRC / IE1 x re1
= 0.253mA x 20k / 0.25mA x 104 Ohm
= 5.06v / 26mV
= 0.19 v

Av2 = 12v / IE2 x re2
= 12v / 3.15mA x 8.25 Ohm
= 12v / 25.99mV
= 0.462 v
 

Harald Kapp

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IE1 = (VCC – VCE) / (R1 + R2 + R3)
Why should this be so?
What about IB2 which also flows through R1?

Let assume VCE Q1 = 6.0v
What makes you think so? Why not 5V or 7V?

Some equations to work with:
IE1=(1+beta)*IB1
IE2=(1+beta)*IB2
voltage across R5: (IE2-IB1)*R5=IB1*R6+VBEQ1+IE1*(R2+R3)
...
I hope that helps.
 
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